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Question Number 148914 by naka3546 last updated on 01/Aug/21

Solve  the  equation     2^x  + x = 11  with  Omega  Function .

$${Solve}\:\:{the}\:\:{equation} \\ $$$$\:\:\:\mathrm{2}^{{x}} \:+\:{x}\:=\:\mathrm{11} \\ $$$${with}\:\:{Omega}\:\:{Function}\:. \\ $$

Commented by naka3546 last updated on 01/Aug/21

thank you, sir.

$${thank}\:{you},\:{sir}. \\ $$

Commented by mr W last updated on 01/Aug/21

2^(11) =(11−x)2^(11−x)   2^(11) ln 2=(11−x)ln 2e^((11−x)ln 2)   (11−x)ln 2=W(2^(11) ln 2)  x=11−((W(2^(11) ln 2))/(ln 2))=3

$$\mathrm{2}^{\mathrm{11}} =\left(\mathrm{11}−{x}\right)\mathrm{2}^{\mathrm{11}−{x}} \\ $$$$\mathrm{2}^{\mathrm{11}} \mathrm{ln}\:\mathrm{2}=\left(\mathrm{11}−{x}\right)\mathrm{ln}\:\mathrm{2}{e}^{\left(\mathrm{11}−{x}\right)\mathrm{ln}\:\mathrm{2}} \\ $$$$\left(\mathrm{11}−{x}\right)\mathrm{ln}\:\mathrm{2}={W}\left(\mathrm{2}^{\mathrm{11}} \mathrm{ln}\:\mathrm{2}\right) \\ $$$${x}=\mathrm{11}−\frac{{W}\left(\mathrm{2}^{\mathrm{11}} \mathrm{ln}\:\mathrm{2}\right)}{\mathrm{ln}\:\mathrm{2}}=\mathrm{3} \\ $$

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